Fix a language $\mathcal{L}$ of first-order set theory. For this question, we can assume that $\mathcal{L}$ is the language described in Chapter 1 of “An introduction to set theory” [William A. R. Weiss | October 2, 2008].
Assuming that the complexity of a formula is not restricted and every formula has a finite length, construct an infinite list $Z_{\mathcal{L}}$ of all syntactically valid statements, so that every such statement is assigned an unique natural number (i.e. enumerate all syntactically valid statements).
Consider an infinite binary sequence $s$. We can assume that any such sequence corresponds to a set of axioms of a particular set theory $T$: an $i$-th bit of $s$ is non-zero if and only if an $i$-th statement of $Z_{\mathcal{L}}$ is an axiom of $T$. Let $t(s)$ denote a theory encoded by $s$. (For example, there will be some $s$ such that $t(s) = \text{ZFC}$.)
Then $f(s) = \alpha+1$ if and only if there exists the smallest ordinal $\alpha$ such that $V_{\alpha} \models t(s)$; otherwise, $f(s) = 0$. For example, if $t(s)=\text{ZFC},$ then $f(s)$ is equal to the successor of the initial ordinal of the smallest worldly cardinal.
Note that the phrase “any binary sequence corresponds to a set of axioms of a particular set theory $T$” does not make any assumptions about $T$: the theory may be empty (if all bits of $s$ are zero) or inconsistent. That is, almost all sequences will not correspond to a consistent theory, but we are not interested in such sequences: for any such sequence $s$ we will have $f(s) = 0$.
The ordinal $\beta_{\mathcal{L}}$ is defined as follows: $$\beta_{\mathcal{L}} = \sup \{ f(s) : s \in {2^\omega }\}.$$
Here “$s \in {2^\omega}$” means that we take into account all infinite binary sequences (subsets of $\omega$).
Question: is $\beta_{\mathcal{L}}$ a well-defined ordinal? If no, why? If yes, how large is $\beta_{\mathcal{L}}$ in the hierarchy of large cardinals?